# Find an eigenvalue decomposition of a hermitian matrix

Suppose A $\in \mathbb C^{m\times m}$ has an $SVD: A = U\sum V^*$. Find an eigenvalue decomposition form of the $2m \times 2m$ hermitian matrix $$B=\begin{bmatrix} 0&A^* \\ A&0 \end{bmatrix}$$

I cannot get the eigenvalue decomposition form of $B=X\sum X^*$. How to do that?

## 1 Answer

We have $$B=\begin{pmatrix}0&V\\U&0\end{pmatrix}\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}\begin{pmatrix}0&U^*\\V^*&0\end{pmatrix}.$$ and $$W\begin{pmatrix}0&\Sigma\\\Sigma^*&0\end{pmatrix}W^*=2\begin{pmatrix}(\Sigma\Sigma^*)^{1/2}&0\\0&-(\Sigma^*\Sigma)^{1/2}\end{pmatrix},$$ where $$W=\begin{pmatrix}(\Sigma^*)^{1/2}&(\Sigma^*)^{-1/2}\\ \Sigma^{1/2}&-\Sigma^{-1/2}\end{pmatrix}.$$ It remains to compute $W^{-1}$.

• In this form we can get $B=\begin{bmatrix} a&0 \\ 0&b \end{bmatrix}$ But in the problem the B is in the form $B=\begin{bmatrix} 0&a \\ b&0 \end{bmatrix}$ – Gatsby Sep 10 '16 at 20:18
• I have corrected the answer. Sorry for possible misprints, not enough time to check. – Vladimir Sep 10 '16 at 20:53
• How do you get the second row? Is this a trick or just by trial and error? – Gatsby Sep 10 '16 at 22:45
• The eigenvalues of $\begin{pmatrix}0&\Sigma\\\Sigma&0\end{pmatrix}$ are those of $\begin{pmatrix}(\Sigma^*\Sigma)^{1/2}&0\\0&-(\Sigma^*\Sigma)^{1/2}\end{pmatrix}$ (this is easy to check by inspecting the characteristic polynomial, because $\Sigma$ is diagonal). In the one-dimensional case, it is also easy to find $W$ reducing our matrrix to this form. The general case follows readily. – Vladimir Sep 11 '16 at 4:52